# Hardy-Weinberg principle

> Biology · CIE A-Level 9700
> Source: https://www.owlsprep.com/study/cie-9700-u18-hardy-weinberg-principle/

This module covers Hardy-Weinberg assumptions, calculation workflows, equilibrium deviation analysis, and exam-style problem solving for CIE A Level Biology.

**Prerequisites:** [Basic understanding of alleles, genotypes and phenotype expression](https://www.owlsprep.com/study/cie-9700-u16-genetic-inheritance-basics/); Simple probability rules for combined independent events

## Learning objectives

- State the 5 core assumptions of the Hardy-Weinberg principle as required for CIE exams
- Apply the standard Hardy-Weinberg equations to calculate unknown allele and genotype frequencies
- Explain why deviations from Hardy-Weinberg equilibrium indicate active evolutionary change
- Solve structured exam problems involving recessive monogenic disorders in human or animal populations

## Core Assumptions of Hardy-Weinberg Equilibrium

The Hardy-Weinberg principle describes a theoretical non-evolving population where allele frequencies remain constant across successive generations. CIE exam mark schemes require you to state all 5 core assumptions to earn full marks for any related definition question.

**Hardy-Weinberg Equilibrium** — A stable population state where no evolutionary forces act, so allele and genotype frequencies do not change from one generation to the next.

- No random mutations occur at the target gene locus
- Mating between individuals in the population is completely random
- There is no significant gene flow (immigration or emigration) between populations
- No natural selection acts to favour or disfavour any specific genotype
- The population is very large, eliminating random genetic drift effects

> **mnemonic**
>
> Use the mnemonic 'Mating Must Not Go Wrong' to remember the 5 assumptions: Mutation, Mating random, No gene flow, No selection, Large population size.

**Check your understanding**

Test your understanding of the assumptions:

1. Which of the following violates Hardy-Weinberg equilibrium?

   - A large isolated population
   - Non-random mating for a preferred trait
   - No new mutations
   - Equal survival of all genotypes

   *Why:* Non-random mating directly shifts genotype frequencies away from expected equilibrium values.

> **Exam tip:** Never list fewer than 5 assumptions in exam answers, as CIE awards 1 mark per stated condition.

## Hardy-Weinberg Equations and Standard Calculation Workflow

The two core equations form the basis of all Hardy-Weinberg calculations. They only apply to diploid, autosomal genes with two possible alleles, and cannot be used for X-linked traits unless explicitly stated in the question.

$$p + q = 1$$

$$p^2 + 2pq + q^2 = 1$$

**Worked example:** If 1 in 2500 people in a population have a recessive autosomal disorder, calculate the frequency of the recessive allele q.

1. First, identify that the proportion of affected individuals equals the homozygous recessive genotype frequency q²
2. $$q^2 = \frac{1}{2500} = 0.0004$$
3. Take the square root of q² to get q
4. $$q = \sqrt{0.0004} = 0.02$$
5. Use p + q = 1 to find the dominant allele frequency p if required
6. $$p = 1 - 0.02 = 0.98$$

**Check your understanding**

Quick calculation check:

1. If q = 0.1, what is the heterozygous carrier frequency?

   - 0.01
   - 0.18
   - 0.81
   - 0.9

   *Why:* 2pq = 2 * 0.9 * 0.1 = 0.18

> **Exam tip:** Always start calculations from the recessive phenotype value, as you cannot distinguish homozygous dominant and heterozygous individuals from external observation alone.

## Interpreting Deviations from Hardy-Weinberg Equilibrium

If observed genotype frequencies do not match the values predicted by the Hardy-Weinberg equations, at least one of the 5 core assumptions has been violated. This deviation is direct evidence that the population is undergoing evolutionary change.

- Higher than expected q² frequency: No selection against the recessive homozygote, or new mutation increasing recessive allele count
- Lower than expected q² frequency: Strong selection against the recessive homozygote, e.g. lethal recessive disorder
- Higher than expected heterozygote 2pq frequency: Heterozygote advantage, e.g. sickle cell trait malaria resistance
- Lower than expected heterozygote 2pq frequency: Inbreeding, which increases homozygote frequency across the genome

**Worked example:** Observed heterozygote frequency for the sickle cell gene in a malaria region is 0.32, but expected Hardy-Weinberg value is 0.18. Explain this deviation.

1. The deviation shows more heterozygotes are present than predicted
2. This is caused by heterozygote advantage: individuals carrying one sickle cell allele have higher resistance to malaria
3. This violates the no selection assumption of Hardy-Weinberg equilibrium

## Exam Phrasing and Mark Scheme Best Practices

**Exam command terms**

CIE uses specific command terms for Hardy-Weinberg questions, each with defined mark requirements:

- **State** — List the 5 assumptions with no extra explanation required

- **Calculate** — Show full working for every step to earn method marks even if your final answer is wrong

- **Explain** — Link deviation from equilibrium directly to a violated Hardy-Weinberg assumption

## Common pitfalls

- **Wrong:** Using the total dominant phenotype frequency directly as p
  - Why it fails: Dominant phenotypes include both p² and 2pq genotypes, so you cannot extract p from this combined value
  - Correct: Always start calculations from the recessive homozygote q² value to derive q first
- **Wrong:** Omitting one or more of the 5 core Hardy-Weinberg assumptions in written answers
  - Why it fails: CIE mark schemes award 1 mark per stated assumption, so partial lists lose easy marks
  - Correct: Memorise and write all 5 assumptions every time you are asked for Hardy-Weinberg conditions
- **Wrong:** Calculating heterozygote frequency as p*q instead of 2*p*q
  - Why it fails: This ignores the two possible allele combinations from parental gametes (dominant from mother + recessive from father, and vice versa)
  - Correct: Double check the heterozygote formula multiplies the product of p and q by 2
- **Wrong:** Rounding intermediate calculation values to 2 significant figures
  - Why it fails: Small rounding errors accumulate and produce final answers outside the CIE accepted tolerance range
  - Correct: Keep 3+ significant figures for all intermediate values, only round the final answer to 2 or 3 s.f.
- **Wrong:** Applying Hardy-Weinberg equations to X-linked traits without adjusting the formula
  - Why it fails: The standard p² + 2pq + q² =1 formula only applies to diploid autosomal loci
  - Correct: Only use the standard equations for autosomal genes, and follow any special instructions given for X-linked cases

## Cheatsheet

| Quantity | Formula | Exam Notes |
| --- | --- | --- |
| Dominant allele frequency | $p$ | Calculated as $1-q$ |
| Recessive allele frequency | $q$ | Square root of recessive phenotype proportion |
| Homozygous dominant frequency | $p^2$ | Cannot be counted from standard phenotypes |
| Heterozygous carrier frequency | $2pq$ | Most commonly asked calculation |
| Homozygous recessive frequency | $q^2$ | Equals proportion of affected individuals |

## What's next

Mastery of the Hardy-Weinberg principle unlocks 6-8 mark calculation questions that appear in almost every CIE A Level Biology Paper 4, and forms the quantitative foundation for understanding how natural selection shifts allele frequencies over generations. You can now connect these theoretical calculations to real-world evolutionary scenarios, from antibiotic resistance in bacterial populations to directional selection for camouflage traits in wild animal groups. This skill will also help you interpret population genetic data in practical exam assessments.

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